ch.4 torsion liu jiemin @sjzu 2008. ch.4 torsion liu jiemin @sjzu 2008 chapter4 torsion 4.1...

LIU Jiemin @SJZU 2008CH.4 TORSION


Chapter4 Torsion Chapter4 Torsion

4.1 Introduction

4.2 Twisting couples and torque

4.3 Torsional stresses in a circular shaft

4.4 Torsional strength of circular shaft

4.5 Torsional deformation of circular shaft, Torsional stiffne

ss condition

4.6 Torsion of noncircular shafts

4.7 Torsion of thin-walled shafts


Shaft ( 轴 ) ： Members subjected mainly to torsion

Torsion: 外力的合力为一力偶，其作用面与轴线垂直 , 变形的特点：两横截面发生绕轴线的相对转动。Circular torsion:

① Torque calculation

② Torsional shear stress and strength

③ Torsional deformation and stiffness

M

F

F

4.1 Introduction


4.1 Introduction


B

A

剪应变：直角的改变量。

M

M

扭转角：任意两截面绕轴线转动而发生的角位移。

O

x

4.1 Introduction



4.1 Introduction





ss condition




4.2.1 Twisting couple exerted to transmission shaft

Power P inputing to a transmission shaft =Twistingcouple M times angle velosityω MP

n

P9549

n

P

2

1060M

m/sN60

n2Mm/sN10P

m/sN60

n2MkWP

r/min

kW3

mN

3

}{

}{}{

min/r

HFmN n

P7024M

m/sN60

n2Mm/s735NP

m/sN60

n2MHFP

}{

}{}{

4.2 External moment

Torque&Torque Diagram


2. Symble of T ： Torque is povitive when the torque vector leaves the cross-section.

4.2.2 Torque and torque diagram

MM

M T

MT

MT

mx

0

0

x

3. Determination of T by (simple) section method

1. Torque ： Moment on cross-sction ， T

n

n

x( )x iT M F

4.2 External moment,Torque&Torque Diagram


4. Torqur diagram ： Curve showing torque veriation with cross-section position

Usage Determine critical section and strength caculation easily

x

T

2T

2T TT

T



Ex.1 The transmission shaaft. Knowing ： n =300r/min ， active wheel PA = 500 kW ， Passive PB=150kW ， PC=150kW ， PD=200kW ， Plot the torque diagram.

nB C A D

PB PC PA PD

①Twisting couples

m)15.9(kN 300

5009.55

n

P.M A

A 559

m)(kN .4300

1509.55

n

P.MM B

CB 78559

m)(kN .300

2009.55

n

P.M D

D 376559

Solution ：



②Find Torques

mkN.MT

MT , m

B

BC1

784

00

1

1

mkN...

MMT CB

5697847842

mkN.MT D3 376

1

1

2

2n

B C A D

MB MC MA MD

3

3

By the same token ：

③Torque diagram x

T (kNm)

4.78

9.56

6.37

––

mkN .Tmax

569




4.1 Introduction





ss condition




1. Cross-section remains plane and undistorted, only rotates differential amounts about x-axle ；

2. Non axial deformation

4.3.1 Observation of experiment


Strength condition


4.3.2 Shearing stress on cross-section:

1. Geometry relation：

xd

d

xd

GGtan

1

xd

d

(R)

2. Physical relation：

xd

dGG

GHooke’law ：


Strength condition


3. Static mechanics relation：

O

dA

Adxd

dG

Adxd

dG

AdT

A

A

A

2

2

AdI Ap2set

xd

dGI T p

pGI

T

xd

d

xd

dG

pI

T


Strength condition


① Only valid for Isotropy, linear elasticity materials under small deformation condition

② T-Torque on cross-section ， obtained from section method

-Distance of the point where shearing stress τis calculated from the center of the circular section.

Ip-Polar inertia moment ， a geometry amount

pI

T

4. Discussion:


Strength condition


R/I

T

I

TR

ppmax

4.3.3 Maximum torsional stress

R/IW pp

pmax W

T

Section modulus in torsion

AdI Ap2

Popar inertia moment


Strength condition


Unit ： mm4 ， m4.

44

2

0

2

2

1032

2

D.D

d

AdID

Ap

For circular section ：

D

d

O

dD

D/d

DI p

)( 4

4

132

For circular ring section ：

16

DW

3

p

)( 41


Strength condition


④ Stress distribution

（ Circular section ）（ Circular ring section）

Using circular ring section:

enhancing strength, save material, light in weight


Strength condition


Thin-walled cylinder：

010

1R （ R0 ： Average radius ）

dA

R0

τdistributes uniformly

TR R

TRAd A

00

0

2

A

T

R

T

020 22

A0 ： Area corresponding to R

0

4.3.4 Torsional stress in thin-walled cylinder


Strength condition


a

c d

b

´

´

dA

R0

)(

12

EG

薄壁圆筒扭转试验可建立剪切胡克定律

G


Strength condition



4.1 Introduction





ss condition




4.4.1 Ultimate stress in in torsion

Mild steel:τs

Cast iron ： fracture along about 45 with axle :τb

Maximum shearing stress when yielding or fracture

45

u

4.4 Strength condition and rational design


4.4.2 Strength calculation of circular shaft

Strength condition ：

Circular shaft with equal section ：

][)( maxP

max W

T

][W

T

p

max

Three types of strength condition ：

① Check ：

② Design section ：

③ Find allowable load ：

][ p

maxmax W

T

][max

p

TW

][pmax WT

nu ][Allowable stress



Ex.4.3 The variable shaft with th same average radias R0, length l=1m, subjected to uniform moment m=3500Nm/m. Chech the strength of the shaft. []=50MPa,knowing that:

m/mN 3500 3500m m/mN

l/2 l/2

x

CA

B

0 AB BCR =50mm, δ =5mm,δ =4mm



For critical section A

② Check strength

][

6MPa.445502

10003500

R2

ml

W

T

2

AB20p

max1

m/mN 3500 3500m m/mN

l/2 l/2

x

CA

B

T

x

ml

( )AM = ml, T = m l - x

Solution ：① Analysis of torque

For critical section B

B2 2

p 0 BC

27 9MPa [ ]W 2 R

T ml /.

2



例 4.2 功率为 150kW ，转速为 15.4 转 / 秒的电动机转子轴如图，许用切应力 []=30M Pa, 试校核其强度。

nNmTBC 2

103

m)(kN551

m)(N4151432

10150 3

.

..

Tm

解：①求扭矩及扭矩图

② 计算并校核剪应力强度

③ 此轴满足强度要求。

D3 =135 D2=75 D1=70A B C

m m

x

][MPa2316070

105513

3

max

.

.WT

t




4.1 Introduction





ss condition




4.5.1 Torsional deformation

From:

pGIT

x

dd

Knowing ： twisting angle between two cross-sections with length l:

l

p

xdGI

T d

0

pGI

Tl

IG

lT

ipi

ii

4.5 Torsional deformation and stiffness condition


4.5.2 Unit torsional angle (rad/m) dd

pGIT

x

/m)( 180 dd

pGIT

x

or

4.5.3 Stiffness condition (rad/m) max pGI

T

/m)( 180 max pGI

T or

GIp: Measure capability of the section resisting torsional

deformation, called as Torsion stiffness of section

[ ]: Allowable unit torsional angle 。



Three types of stiffness calculation ：

① Strength check：

② Design section ：

③ Fing allowable load：

max

] [

max

GT

I p

] [ max

pGIT

Sometimes ， select materials by the condition



Ex.4 The transmission shaft. n = 500 r / min ， Inputting

P1 = 400kW, Outputting P2 = 150kW, P3 = 250kW.

Knowing ： G=80GPa ， [ ]=70M Pa ， [ ]=1º/m ，Determine ： ① d1 for portion AB and d2 for portion BC ？ ②If the diameter of shaft is same ， how to select d ？ ③How to arrange rationally active wheel and passive wheel ？

Solution ：⑴ Plot torque D 500 400

P

1

P3P2

A CB

T x

7.644.775

(kNm)

11

22

32

P=7.64kN m

P=2.865kN m

P=4.775kN m

m .n

m .n

m .n

9 55

9 55

9 55



][ Td

W p 16

3

3

16 4775000=70.3mm

3.14 70

Td

32

16

] [ 32 4

GTdI p

3

2

16 7640000N mm=82.2mm

3.14 70N/mm

Td

31

16

⑶ From Stiff.C ：

⑵ From S.C. ：

6

4 2 32 3

32 7.64 10 N mm 18086.4mm

N 1 [ ] 3.14 80 10mm 10 mm

Td

G

41

32

6

42

32 4.775 10 180=76.68mm

[ ] 3.14 80 1

Td

G

42

32



轴上的绝对值最大的扭矩越小越合理，所以， 1 轮和 2 轮应该换位。换位后 , 轴的扭矩如图所示 , 此时 , 轴的最大直径才为 75mm 。

T

x

4.21

(kNm)2.814

1 2 mm mmd , d 87 77

② 全轴选同一直径时 mmd d 1 87

①

T x

7.0244.21

(kNm)

Result:

③



例 5 长为 L=2m 的圆杆受均布力偶 m=20Nm/m 的作用，如图，若杆的内外径之比为 =0.8 ，外径 D=0.0226m ， G=80GPa ，试求固端反力偶。

解：①杆的受力图如图示，这是一次超静定问题。平衡方程为：

02 BA mmm



② 几何方程——变形协调方程 0BA

③ 综合物理方程与几何方程，得补充方程：

0402202

00

P

A

P

AL

PBA GI

mdx

GIxm

dxGIT

mN 20 Am

④ 由平衡方程和补充方程得：

另 : 此题可由对称性直接求得结果。

mN 20 Bm


(rad/m) dd

pGIT

x

l/2

l/2

l/2m/mN 3500 3500m m/mN

l/2 l/2

x

CA

B


例 3.4 组合机床主轴箱内第四轴。动力由 5 轴经Ⅲ齿轮输入到 4 轴，齿轮Ⅱ和 Ⅳ带动轴 1 、 2 和 3 。轴 1 、 2 钻孔，共消耗功率 0.755kW; 轴 3 扩孔，消耗功率 2.98kW 。若轴 4转速为 183.5r/min,G=80GPa,[τ]=40 MPa,[θ]=1.5°/m 。试设计轴的直径。

4 ⅣⅡ

5

1 ， 2 3

Ⅲ



4 ⅣⅡ

5

Ⅲ

1 ， 2 3mⅢ

(a)

mⅡmⅣ(b)

39.3Nm

155 Nm

-

+

扭矩图(c)



(1) Calculation of externall torque: m Ⅱ=9549N Ⅱ /n=9549×0.735/183.5=39.3 Nm m Ⅳ=9549N Ⅳ /n=9549×2.98/183.5=155 Nm m Ⅲ= m Ⅱ+ m Ⅳ=39.3+155=194.3 Nm(2)Design of the shaft diameter Tmax = 155 Nm From strength condition: τ max =Tmax /Wp=16Tmax/(πD3 ) ≤[τ] D1≥0.0272m

39.3Nm

155 Nm

-

+

扭矩图


Solution:


From stiffness condition:

θ max =Tmax ×180/(GIp×π) ≤[θ]

D2 ≥0.0297m

D>D2 =30mm

Finally, The diameter of the shaft is determined as 30 mm. It can be seen that the design of the shaft is controlled by stiffness condition.



φ

m

轴受扭而转

薄壁筒不受扭

例 3 。 5 （ 107 页）轴受扭矩产生扭转角 φ 。这时把薄壁筒与轴的凸缘焊接在一起，然后解除扭矩 m 。设轴和筒的抗扭刚度分别是 G1Ip1 和 G2Ip2 试求轴内和筒内的扭矩。



φ1

φ2

φ

T2

T1



Solution: (1) Equilibrium equation: T1-T2=0 (1) (2) Compliant condition: φ1 + φ2 = φ (2)

(3) Physical relationship: φ =ml/(G1Ip1) φ1 =T1l/(G1Ip1) (3) φ2 =T2l/(G2Ip2)

From Eqs.(1)-(3),

T1=T2=mG2Ip2 /(G1Ip1+ G2Ip2)



1. 点 M 的应力单元体如图 (b) ：

(a)

M

(b)

´́ ´́

(c)

2. 斜截面上的应力；

取分离体如图 (d)

：

(d)

´́

x



(d)

´́

x

n

t

转角规定：

轴正向转至截面外法线逆时针：为“ +”

顺时针：为“–”由平衡方程：

0)cossind()sincosd(d ; 0 AAAFn

0)sinsind()coscosd(d ; 0 AAAFt

解得： 2cos ; 2sin



2cos ; 2sin 分析：

当 = 0° 时，

max00 , 0

当 = 45° 时，

0 , 45min45

当 = – 45° 时，

0 , 45max45

当 = 90° 时，

max9090 , 0

´́

45°

由此可见：圆轴扭转时，在横截面和纵截面上的剪应力为最大值；在方向角 = 45 的斜截面上作用有最大压应力和最大拉应力。根据这一结论，就可解释前述的破坏现象。


ch.4 torsion liu jiemin @sjzu 2008. ch.4 torsion liu jiemin @sjzu 2008 chapter4 torsion 4.1...

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